# Why does tetrachlorocuprate(II) form in aqueous solution even though water should be the stronger ligand? [closed]

In aqueous solution, $\ce{Cu^{2+}}$ forms the $\ce{[Cu(H2O)6]^2+}$ complex. Given that water is a stronger ligand than $\ce{Cl-}$, though, why does why does the $\ce{[CuCl4]^2-}$ complex form upon addition of chloride ions?

• Upon addition of more chloride ions to what exactly? o_o Feb 15 '17 at 15:33
• To that same aqueous solution in which [Cu(H2O)6]2+ ions are present. Feb 16 '17 at 17:55

[Water ligands] being stronger than $\ce{Cl-}$ ligands will form a $\ce{[Cu(H2O)6]^2+}$ complex.
Which ligand forms bonds of which strength to which metal is only very generally determinable; and in general you would always denote binding affinities in a $1:1$ manner: ligand $\ce{X}$ forms strong bonds to metal $\ce{Y}$. So sulphide ions will form strong dative bonds to mercury(II) ions but not-so-strong bonds to titanium(IV) ions. Likewise, palladium(II) if presented with both phosphane and amino ligands will preferentially bind to phosphanes while copper(I) will preferentially bind to amino ligands.
To answer the actual question you posed: $\ce{[CuCl4]^2-}$ will form because it is the predominant species under a given chloride ion concentration. This can be backrationalised by looking at formation constants $K_1$ to $K_4$. However, aside from quantum chemical calculations there is no way to properly rationalise it a priori.